A secant line intersects the curve $y=x^2+5x$ at two points with $x$ -coordinates $2$ and $2+h$, where $h\neq0$. What is the slope of the secant line in terms of $h$ ? Your answer must be fully expanded and simplified.
Answer: We are given that the secant line intersects the curve at $x=2$ and $x=2+h$. Since these points are on the curve $y=x^2+5x$, we know that their $y$ -values are $y=14$ and $y=(2+h)^2+5(2+h)$, correspondingly. To summarize this part, we know that the secant line passes through the points $(2,14)$ and $(2+h,(2+h)^2+5(2+h))$. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{(2+h)^2+5(2+h)-14}{2+h-2} \\\\ &=\dfrac{(2+h)^2+5(2+h)-14}{h} \end{aligned}$ We can now simplify the expression we obtained. $\begin{aligned} &\phantom{=}\dfrac{(2+h)^2+5(2+h)-14}{h} \\\\ &=\dfrac{4+4h+h^2+10+5h-14}{h} \\\\ &=\dfrac{h^2+9h}{h} \\\\ &=\dfrac{h(h+9)}{h} \\\\ &=h+9\text{, for }h\neq 0 \end{aligned}$ Since we are given that $h\neq 0$, we can conclude that the slope of the secant line is $h+9$.